What is the value of the following logarithm? $\log_{3} \left(\dfrac{1}{9}\right)$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $3^{y} = \dfrac{1}{9}$ In this case, $3^{-2} = \dfrac{1}{9}$, so $\log_{3} \left(\dfrac{1}{9}\right) = -2$.